# Chemical Kinetics: Elementary Reactions

Shaun Williams, PhD

## Reaction Rates

### Reaction Rate

• Chemical kinetics is the study of chemical reaction rates.
• In order to understand kinetics, there are some background equations that we need.
• These equations will be fully developed later in the course.

### The Collision Frequency

• $$\gamma = \rho \sigma \left< v_{AA} \right>$$
• The effective cross-sectional area of the colliding molecules is $$\sigma_{AB}=\frac{1}{4}\left( \sigma_A + 2\sqrt{\sigma_A \sigma_B} +\sigma_B \right)$$
• The average relative speed is $$\left< v_{AB} \right> = \sqrt{\frac{8k_\mathrm{B}T}{\pi \mu}}$$
• The reduced mass is $$\mu = \frac{m_A m_B}{m_A+m_B}$$

### Fick's First Law

• Some reaction rates are limited by diffusion
• In these cases, we will use Fick's first law to help quantify the dynamics.
• $$J(Z_0) = -D\left. \left( \frac{d\rho}{dZ} \right) \right|_{Z_0}$$

### Equilibrium Constant

• The equilibrium constant can be written in terms of partition function (a lot more on that later in the course) $$K_{\mathrm{eq}} = \frac{Q_{\mathrm{products}}}{Q_{\mathrm{reactants}}}$$
• We can also write it in terms of the standard state Gibbs' Free energy $$\Delta_{\mathrm{rxn}} G^\ominus$$ $$K_\mathrm{eq} = e^{-\frac{\Delta_\mathrm{rxn}G^\ominus}{RT}}$$
• The Gibbs' Free energy can be broken down as $$\Delta G = \Delta H - T \Delta S$$
• The enthalpy for any process is given as $$H = E + PV$$

### Reaction Rates

• Chemical kinetics is the study of chemical reaction rates.
• In kinetics, both microscopic and macroscopic parameters of the system are crucial to understanding chemical behavior

### Microscopic Parameter: Orientation of Reactants

• Consider the reaction $$\mathrm{Cl(g) + HF(g) \rightarrow HCl(g) + F(g)}$$
• For this reaction $$\Delta_\mathrm{rxn}H^\ominus = 165.198+173.779-186.908-158.754 = -6.685\,\frac{kJ}{mol}$$
• This reaction is less probable if the chlorine attacks from the F end of the HF molecule.
• This reaction is even less probable if the fluorine atom is replaced wiht much larger functional groups. $$\mathrm{Cl(g) + (CF_3)_3CH(g) \rightarrow HCl(g) + (CF_3)_3C(g)}$$
• Because of these steric considerations we should expect that the first reaction will be much more probable than the second reaction
• This means that we should expect the reaction rate of the first reaction to be faster than the reaction rate of the second reaction.

### Microscopic Parameter: Energy Barrier to Reaction

• During a chemical reaction, the potential energy of the system tends to increase.
• The total energy of the system, however, is constant.
• Basically, the reacting system is exchanging kinetic energy (vibrational and rotational) for potential energy.
• A chemical reaction must overcome the activation energy in order to react.
• The activation energy is the peak in potential energy between the reactants and products.
• The height of this barrier depends on the molecular structure and quantum mechanics.

### Macroscopic Parameters

• The rate at which we observe changes in a system depends on the system's thermodynamic parameters.

#### Concentrations

• The language of chemical kinetics grew out of the study of reactions in solutions.
• We tend to represent the amount of each molecule in formulas using molarity.
• Even in gas phase reactions we describe the concentration using the number density is particles per $$\mathrm{cm}^{-3}$$

#### Temperature

• The rate constant is a function of temperature (we will later determine this dependence)

### Reaction Velocities

• We can define a reaction velocity as $$v=\frac{1}{\nu_i}\frac{d\left[ i \right]}{dt} = \frac{1}{V\nu_i}\frac{dn_i}{dt}$$
• $$\left[ i \right]$$ - molar concentation of product or reactant $$i$$
• $$\nu_i$$ - the stoichiometric coefficient of $$i$$ in the reaction
• $$n_i$$ - the number of moles of component $$i$$ contained in volume $$V$$
• The use of the stoiciometric coefficient in the equation is to ensure that the same reaction rate is found regardless of what component is being considered.
• For the reaction $$\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$$ $$-\frac{1}{2}\frac{d\left[ \mathrm{H_2}\right]}{dt} = -\frac{d\left[ \mathrm{O_2}\right]}{dt} = \frac{1}{2}\frac{d\left[ \mathrm{H_2O}\right]}{dt}$$

### Rate Laws and Complications

• Often, the reaction velocity is written in terms of a rate law
• The rate law is a power law in the reactant concentrations
• The rate law includes a concentration-independent coefficient called the rate constant, $$k$$
• An example of a generic rate law $$v=k\left[ A\right]^x\left[ B\right]^y\dots$$
• The reaction order is the sum of the exponents, $$x+y+\dots$$

### Elementary vs Net Reactions

• A net rreaction is a reaction written to show only the initial and final chemical components, for example $$\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$$
• This reaction actually occurs in a series of steps \begin{align} \mathrm{H_2 + O_2} & \rightarrow \mathrm{H+HO_2} \\ \mathrm{HO_2 + H_2} & \rightarrow \mathrm{H_2O + OH} \\ \mathrm{H + OH + M} & \rightarrow \mathrm{H_2O + M} \end{align} The $$M$$ in the last step is shorthand for any particle that participates in the reaction without being changed
• Each of these individual steps is called an elementary step.
• Elementary steps require only a single collision or decay.

### Reaction Mechanisms

• The reaction mechanism is the series of elementary reaction that lead to a net reaction equation
• On the previous slide, if we add up the three elementary steps, we get the net equation
• In this chapter we will focus on elementary reactions

### Types of Elementary Reactions: Unimolecular

• If the initial reactant is unstable $$\mathrm{XeCl \rightarrow Xe + Cl}$$
• If the reactant molecule has been previous energized (through collisions or laser) $$\mathrm{C_2H_4O_2\,(dioxetane) \rightarrow 2H_2CO}$$

### Types of Elementary Reactions: Bimolecular and Termolecular

• A bimolecular elementary reaction involves the collision between two reactants $$\mathrm{Cl + HCl \rightarrow Cl_2 + H}$$
• A termolecular elementary reaction involves the collision between three reactants $$\mathrm{He + H + Cl \rightarrow He + HCl}$$
• It is important to note that the probability of three reactants colliding simultaneously is very small compared to bimolecular collisions
• Elementary reactions involving more than three reactants have not been found.

## Simple Collision Theory

### Simple Collision Theory

• We now what to fold together the parameters to predict the value of the rate constant, $$k$$
• For our simplest model of chemical reactions, we fix the temperature and assume the following
1. The reaction is bimolecular, $$\mathrm{A + B} \xrightarrow{k} \mathrm{products}$$
2. The reactants must be close enough for intermolecular forces to be important before the reaction can take place.
3. A single distribution function describes the probability of the reaction occuring
4. There will be some activation barrier to overcome

### Derivation Summary: Simple Collision Theory

• First we count the number of collisions per unit time
• Multiply that by the fraction of those collision with enough kinetic energy
• this bring in the canonical factor $$e^{-\frac{E_a}{RT}}$$ (later in the course)
• Bringing all this together we arrive at a version of the empirical Arrhenius equation $$k_\mathrm{Arr}=Ae^{-\frac{E_a}{RT}}$$ namely $$k_\mathrm{SCT} = \sigma_\mathrm{AB} \sqrt{\frac{8k_\mathrm{B}T}{\pi \mu}} \mathcal{N}_\mathrm{A} p e^{-\frac{E_a}{RT}}$$ where $$p$$ is a steric factor.

### Behavior of the Arrhenius Rate Constant

• Note: When $$RT=E_a$$, $$k_\mathrm{Arr}$$ is about at one-third of its maximum value.

### Example 13.1

Estimate the value of $$A$$ in $$\mathrm{cm^3\,mol^{-1}\,s^{-1}}$$ for the reaction between oxygen atom and molecular nitrogen, $$\mathrm{O(g) + N_2(g)}$$, at $$298\,\mathrm{K}$$. Let the collision cross-section be $$30\, \AA^2$$ for oxygen and $$37\, \AA^2$$ for $$\mathrm{N_2}$$.

### Using the Arrhenius Equation

• We would like to be able to measure the value of $$k$$ at various temperatures in the lab and determine the value of $$E_a$$
• To do this, we first need to rearrange the Arrhenius equation

\begin{align} k & = A \exp \left( -\frac{E_a}{RT} \right) \\ \frac{k}{A} & = \exp \left( -\frac{E_a}{RT} \right) \\ \ln \frac{k}{A} & = -\frac{E_a}{RT} \end{align}

### First of Two Results in the Derivation

\begin{align} \frac{d}{dT} \left( \ln \frac{k}{A} \right) & = -\frac{d}{dT} \frac{E_a}{RT} \\ \frac{d}{dT} \left( \ln k - \ln A \right) & = \frac{d}{dT} \ln k = -\left( -\frac{E_a}{RT^2} \right) = \frac{E_a}{RT^2} \\ E_a & \equiv RT^2 \frac{d\ln k}{dT} \end{align} This result is great, but experimentally, we needs a more useful approach.

### First of Two Results in the Derivation

Let's try a more useful approach \begin{align} \ln \frac{k}{A} & = -\frac{E_a}{RT} \\ \ln k - \ln A & = -\frac{E_a}{RT} \\ \ln k & = -\frac{E_a}{R}\frac{1}{T} + \ln A \end{align} This is now linear. A plot of $$\ln k$$ versus $$\frac{1}{T}$$ will provide a straight line with a slope of $$-\frac{E_a}{R}$$ and a y-intercept of $$\ln A$$.

### Example 13.2

For the reaction $$2NOBr \rightarrow 2NO + Br_2$$, calculate $$k$$ at $$298\,\mathrm{K}$$ assuming only the Arrhenius equation (constants given in Table 13.1 of the book). Then predict the values of $$k$$ at $$400\,\mathrm{K}$$ if (a) the Arrhenius equation continues to hold and (b) if the Arrhenius value at $$298\,\mathrm{K}$$ is correct but $$A$$ is actually of the form $$A' T^\frac{1}{2}$$.

## Transition State Theory

### Transition State Theory

• Many kinetic models take advantage of transition state theory or activated complex theory
• This theory describes a chemical reaction as a process in which reactants combine to form an activated complex which then decays to product $$\mathrm{A+B \rightarrow AB^\ddagger \rightarrow products}$$

### Eyring Equation

• The rate constant equation that we get utilizing transition state theory is called the Eyring equation $$k_\mathrm{TST} = \frac{k_\mathrm{B}T}{Ch}\exp \left( \frac{\Delta S^\ddagger}{R} \right) \exp \left( -\frac{\Delta H^\ddagger}{RT} \right)$$ where $$C$$ is the overall concentration
• Here we have a slightly more complex version than Arrhenius involving more thermodynamic quantities
• Our steric factor has been replaced by a term dependent on the entropy change which makes sense

## Diffusion-Limited Rate Constants

### Diffusion-Limited Rate Constants

• Our equations thus far have only taken into account gas phase diffusion
• In order to correctly model reactions in solutions we need to allow for the diffusion of the reactants in a solvent
• We are going to use Fick's first law to do this $$J=-D\left( \frac{d\rho}{dZ} \right)$$ where $$J$$ is the diffusion-limited flux of molecules per unit area per unit time

### Diffusion-Limited Reaction

• To estimate the diffusion-limited rate constant, we assume a sphere centered on each molecule A such that any molecule B that gets inside the sphere will react.

### Diffusion Rate

• The rate, in molecules per unit time, of B diffusing into any sphere of radius $$r$$ is given the symbol $$\gamma_\mathrm{D}$$ \begin{align} \gamma_\mathrm{D} & = -J_\mathrm{B} \left( 4\pi r^2 \right) \\ & = D_\mathrm{AB} \left( 4\pi r^2 \right) \left( \frac{d\rho_\mathrm{B}}{dr} \right) \end{align}
• Doing some math we arrive at $$\gamma_\mathrm{D} = 4\pi R_\mathrm{AB} D_\mathrm{AB} \rho_\mathrm{B}$$

### Diffusion-Limited Rate Constant

• Using our diffusion rate equation we can derive a rate constant
• This rate constant assumes:
1. Diffusion-limited rate
2. A bimolecular reaction
• The diffusion-limited rate constant is $$k_\mathrm{D}=4\pi R_\mathrm{AB} D_\mathrm{AB}$$

## Rate Laws for Elementary Reactions

### Rate Law for Elementary Reactions

• Both simple collision and transition state theories find that the reaction rate depends solely on the rate constant and the concentrations of the reactants $$\mathrm{elementary\, reaction:}\, v=k\prod_\mathrm{reactants} \left[ \mathrm{i} \right]^{v_\mathrm{i}}$$
• Some examples: \begin{align} \mathrm{A \rightarrow products:}\; & v=k\left[ \mathrm{A} \right] \\ \mathrm{A + B \rightarrow products:}\; & v=k\left[ \mathrm{A} \right] \left[ \mathrm{B} \right] \\ \mathrm{2A \rightarrow products:}\; & v=k\left[ \mathrm{A} \right]^2 \\ \mathrm{A + B + C \rightarrow products:}\; & v=k\left[ \mathrm{A} \right] \left[ \mathrm{B} \right] \left[ \mathrm{C} \right] \\ \end{align}

### Unimolecular Decomposition

• Let's look at a generic unimolecular decomposition reaction, $$A \rightarrow B + C$$
• The rate law for this reaction is $$v = \frac{d\left[\mathrm{B}\right]}{dt} = -\frac{d\left[\mathrm{A}\right]}{dt} = k\left[\mathrm{A}\right]$$
• We would like to come up with an equation that allows us to determine the concentration of a species as a function of time
• To do this, we need to do a little math

### First Order Integrated Rate Law

• Starting with $$-\frac{d\left[\mathrm{A}\right]}{dt} = k\left[\mathrm{A}\right]$$ we find that: \begin{align} -\frac{d\left[\mathrm{A}\right]}{\left[\mathrm{A}\right]} & = k\, dt \\ -\int_{\left[\mathrm{A}\right]_0}^{\left[\mathrm{A}\right]} \frac{d\left[\mathrm{A}\right]'}{\left[\mathrm{A}\right]'} & = k\int_0^t dt \\ -\ln \frac{\left[\mathrm{A}\right]}{\left[\mathrm{A}\right]_0} & = kt \; \therefore \; \left[\mathrm{A}\right] = \left[\mathrm{A}\right]_0 e^{-kt} \\ \mathrm{if\, \left[B\right]_0 = \left[C\right]_0 = 0:}\; \left[\mathrm{B}\right] & = \left[\mathrm{C}\right] = \left[\mathrm{A}\right]_0\left(1-e^{-kt}\right) \end{align}

### Reaction Half-Life

• Once we have the integrated rate law we can determine the reaction half-life
• The half-life, $$t_\bfrac{1}{2}$$, is the time is takes the reaction for $$\left[\mathrm{A}\right] = \frac{1}{2}\left[\mathrm{A}\right]_0$$
• So \begin{align} -\ln \frac{\left[\mathrm{A}\right]}{\left[\mathrm{A}\right]_0} & = kt \\ -\ln \frac{\frac{1}{2}\left[\mathrm{A}\right]_0}{\left[\mathrm{A}\right]_0} & = kt_\bfrac{1}{2} \\ -\ln \frac{1}{2} & = kt_\bfrac{1}{2} \; \therefore \; t_\bfrac{1}{2} = -\frac{1}{k}\ln \frac{1}{2} = \frac{\ln 2}{k} \end{align}

### Example 13.3

Find the integrated rate law for the elementary reaction $$\mathrm{2A\rightarrow B}$$, where $$\mathrm{A}$$ is butadiene and $$\mathrm{B}$$ is vinylcyclohexene. Use this to obtain an expression for the half-life of butadiene in terms of the rate constant $$k$$ and the initial concentration $$\left[ \mathrm{A} \right]_0$$.

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